Double dual is isomorphic to the vector space - problem with injectivity This is my proof that $V$ is isomorphic to $V^{**}$:


When I say: "Since this is valid for any $w\in V^*$, then $v=0$", is it correct? I do not see why. I wrote it "just because", expecting to figure it out later.
 A: What you want is the following lemma:
If $v \in V$ is such that for all $w \in V^*$, one has $w(v) = 0$, then $v = 0$. 
To prove the lemma, it's easier to prove the contrapositive: if $v \neq 0$, we can explicitly build a function $w \in V^*$ that doesn't vanish on $v$. To do this, note that since $v \neq 0$, the set $\{v\}$ is linearly independent, so, by a result you have hopefully already seen in the course, it can be extended to a basis $\{v, v_1, v_2, \ldots, v_{n-1}\}$. 
Now there is a unique $w \in V^*$ with $w(v) = 1$ and $w(v_i) = 0$ for the other basis vectors. (It is uniquely determined from these rules by linearity, since the above set is a basis.) In particular, $w$ doesn't vanish on $v$, as desired.
A: Let $dim(V)=m$ and $\{v_1,v_2...v_m\}$ a basis of $V$.
What you said is indeed true.
Let $x=s_1v_1+s_2v_2+...+s_mv_m$
If $f(x)=0$ forall $f\in V^*$ then $x=0$.
Take the linear functionals $f_1,f_2...f_m \in V$ such that $f_i(v_j)=1$ if $i=j$ and $f_i(v_j)=0$ if $i \neq j$
You can prove that these are indeed linear functionalas and use them to prove that $x=0$
